Also note that angular momentum is not a quantity that stands on its own. It is a manifestation of linear momentum at a distance, just like linear velocity is a manifestation of a rotation at a distance.
For me linear momentum is incorrectly characterized as $\vec{p} = m \vec{v}_C$ (where C indicates the center of mass). At any instant a rigid body is rotating about an axis, and if you use $\vec{c}$ the location of the center of mass relative to the rotation axis then the equations of motion are
$$ \begin{align}
\vec{p} & = m \vec{\omega} \times \vec{c} & \vec{F} &= \frac{\rm d}{{\rm d}t} \vec{p} \\
\vec{L}_C &= I_C \vec{\omega} & \vec{\tau}_C & = \frac{\rm d}{{\rm d}t} \vec{L}_C \\
\vec{L} &= \vec{L}_C + \vec{c} \times \vec{p} & \vec{\tau} &= \vec{\tau}_C + \vec{c} \times \vec{F}
\end{align} $$
So your question is simply if you have a clump of particles, each obeying the first row of the equations above, can you derive second row?
Yes. this is how. Take an infinitesimal mass ${\rm d}m$ and find it's contribution to linear momentum (from first equation above).
$$ {\rm d}\vec{p} = (-\vec{c}\times \vec{\omega})\, {\rm d}m$$
Now use the transformation rule (third row of equations) for the infinitesimal particle
$$ {\rm d} \vec{L} = \vec{c} \times {\rm d} \vec{p} = (-\vec{c}\times \vec{c}\times \vec{\omega})\, {\rm d}m$$
A simple integration yields
$$ \vec{L} = \int (-\vec{c}\times \vec{c}\times \vec{\omega})\, {\rm d}m = \left[ \int (-\vec{c}\times \vec{c}\times )\, {\rm d}m \right] \vec{\omega}$$
But what is inside the integral? Each $[\vec{c}\times]$ is actually a 3×3 skew-symmetric matrix representing the cross product operator $$ \begin{pmatrix} x \\ y \\ z \end{pmatrix}\times = \begin{vmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{vmatrix}$$
Using the position vector $\vec{c} = (c_x,c_y,c_z)$ the contents of the integral are
$$ I = \int \begin{vmatrix}
c_y^2+c_z^2 & -c_x c_y & -c_x c_z \\
-c_x c_y & c_x^2 + c_z^2 & -c_y c_z \\
-c_x c_z & -c_y c_z & c_x^2 + c_y^2
\end{vmatrix}\,{\rm d}m $$
You will recognize the above as the definition of the mass moment of inertia (about the rotation axis) if you look at any dynamics book. Se we have shown that $$ \vec{L} = I \vec{\omega} $$. The reason that we do the equations of motion about the center of mass and not the center of rotation is that in general we know where the COM location is, but not where the COR is. Also in general the mass moment of inertia matrix is defined about the COM. If $I=I_C$ and $\vec{L} = \vec{L}_C$ which gives us the second row.
The torque equation is derived from the transformation $$\vec{\tau} = \vec{c} \times \vec{F} = \vec{c} \times \frac{\rm d}{{\rm d}t} \vec{p} = \frac{\rm d}{{\rm d}t} \vec{L}$$
